I am assuming by "cuboid" you mean a rectangular box (which Wikipedia calls a "rectangular cuboid", the term "cuboid" being more general).
Then you are asking for a solution to the equations
$$
4(l + w + h) = 2(lw + wh + hl) = lwh.
$$
Setting $c = l + w + h \in \mathbb{R}$, we construct the polynomial
$$
(x - l)(x - w)(x - h) = x^3 - c x^2 + 2c x - 4c,
$$
and we wish to know if it
has three positive real roots for any real $c$.
The discriminant of this polynomial is
$$
-432 c^2+112 c^3-12 c^4 = -4c^2(3c^2 - 28c + 108),
$$
which is strictly negative for all $c$, implying that there are two nonreal roots, and no solution to the equations exists for real numbers $l, w, h$.
We have a cuboid with side lengths $x$, $y$, and $z$. Its volume is
$$V = x y z$$
and area is
$$A = 2 x y + 2 x z + 2 y z$$
When the side lengths are measured in centimeters, we are looking for integer side lengths for which
$$\frac{V}{1 \; cm^3} = \frac{A}{1 \; cm^2}$$
and one of the lengths is the largest possible (integer). We are given $x = 3 cm$.
From now on, for simplicity, we can drop the units. Just remember you cannot do unit conversions; we are assuming the above units.
(When you aren't sure if the method you are trying to apply to solve the problem is sane, you can always apply quick dimensional analysis to see if the dimensions match. If they do not, you are comparing apples and oranges, and most likely the method does not apply -- or more likely, you perhaps remember some details wrong. It is useful to know unit conversions, so you don't get tripped up if you have $[kg \, m/s^2]$ on one side, and $[N]$ on the other: they are equivalent. Some units, like specific impulse used to characterize rockets, may have weird units: if propellant weight is used, specific impulse is typically measured in seconds. In dimensional analysis, you work through the equations, but ignoring the numbers (unless you suspect you might be dividing by zero, or by infinity), and only look at the units. So, it's quite simple, and very powerful. It's saved my bacon a lot of times; I'm often wrong, but I sometimes catch myself, when checking my thinking using tools like dimensional analysis.)
The equation (for area and volume being equivalent) can now be written as
$$x y z = 2 ( x y + x z + y z )$$
and substituting $x = 3$,
$$3 y z = 6 y + 6 z + 2 y z$$
Subtracting $2 y z$ from both sides we get
$$y z = 6 (y + z)$$
which is equivalent to
$$y z - 6 y = 6 z$$
and
$$y (z - 6) = 6 z$$
Dividing both sides by $(z - 6)$ we get
$$y = \frac{6 z}{z - 6}$$
Note that due to symmetry, we also have
$$z = \frac{6 y}{y - 6}$$
You can work out the previous three or four steps to verify it if you wish.
Because of the symmetry regarding $y$ and $z$, without loss of generality, we can declare we want
$$z \ge y$$
When only integer solutions are asked for, it usually suffices to look at the general behaviour (derivatives, long-scale changes), and a small number of potential candidate answers near the interesting points indicated by the general behaviour (discontinuities, extrema i.e. where the derivative is zero, and so on). Occasionally the solutions are more complicated -- say, distances to 2D lattice points --, in which case additional work is needed.
If $y = 6$, then $z$ is undefined due to division by zero. If $y = 7$, then $z = 6 \cdot 7 / (7 - 6) = 42$.
For all $y \gt 7$, $z \lt 42$. One good way to verify this is look at the derivative of $6 y / (y - 6)$:
$$\frac{d z}{d y} = \frac{6}{y - 6} - \frac{6 y}{(y - 6)^2} = -\frac{6^2}{(y - 6)^2}$$
Because the derivative is negative for all $y \gt 6$, it means $z$ is monotonically decreasing for $y \gt 6$: the values of $z$ become smaller and smaller as $y$ grows.
Because the largest $z$ is reached when $y = 7$ (then $z = 42$), and $x = 3$ was given, the solution is
$$\begin{cases}
x = 3 \; cm \\
y = 7 \; cm \\
z = 42 \; cm
\end{cases}$$
Best Answer
Let the cuboid have sides $a,b,c > 0$ and constrain $a+b+c=K$ for a constant $K$. Now optimize the choice of $a,b,c$ to maximize the volume.
You will get $a=b=c=K/3$, yielding your result.
You have $c = K - a -b$ and $$V(a,b) = ab(K-a-b) = Kab - a^2 b - ab^2,$$ which we seek to maximize. Note that $$ \begin{split} V_a(a,b) = \frac{\partial V}{\partial a} &= Kb - 2ab - b^2 = b(K-2a-b)\\ V_b(a,b) = \frac{\partial V}{\partial b} &= Ka - 2ab - a^2 = a(K-2b-a)\\ \end{split} $$ so solving $V_a(a,b) = 0 = V_b(a,b)$, the first equation yields $b=0$ (which is the minimum solution of zero volume) and $2a+b=K$, whereas $V_b(a,b)=0$ yields $2b+a = K$. Hence you have $$ 2a+b = K\\ a + 2b = K, $$ which has a unique solution at $a=b=K/3$.